QFT: Computing S-Operator to 1st Order in Coupling Constant lambda

[WarnK]

Homework Statement

Compute the S-operator to first order in the coupling constant lambda.

Homework Equations

The given Lagrangian density is
$$L = : \frac{1}{2} (\partial_{\mu} \phi)^2 - \frac{1}{2}m^2\phi^2 + \frac{1}{2}\frac{\lambda}{4!}\phi^4 :$$
where phi is a scalar field.

The Attempt at a Solution

S = 1+iT
and I want to calculate iT to first order, which I guess is
$$<0|T\big( -i\int d^4x \frac{\lambda}{4!}\phi(x)^4 \big)|0>$$
using Wick's theorem, how is this anything except zero? Or I'm I missing something?

 

[kdv]

Homework Statement

Compute the S-operator to first order in the coupling constant lambda.

Homework Equations

The given Lagrangian density is
$$L = : \frac{1}{2} (\partial_{\mu} \phi)^2 - \frac{1}{2}m^2\phi^2 + \frac{1}{2}\frac{\lambda}{4!}\phi^4 :$$
where phi is a scalar field.

The Attempt at a Solution

S = 1+iT
and I want to calculate iT to first order, which I guess is
$$<0|T\big( -i\int d^4x \frac{\lambda}{4!}\phi(x)^4 \big)|0>$$
using Wick's theorem, how is this anything except zero? Or I'm I missing something?

Why do you think this should give zero?Wick's theorem tells you to sum over all the possible contractions so you can contract the first field with the second and the third with the fourth or the first with the third and the second with the fourth or the first with the fourth and the second with the third.

 

[olgranpappy]

how is this anything except zero?

each phi is a linear combination of creation and annihilation terms, so it's not too hard to believe that you can get some non-zero piece out of <phi^4>. To say something quantitative you want to write the time ordered product in terms of normal ordered products and contractions... which is done using wick's theorem.

 

[WarnK]

$$<0|T\big( -i\int d^4x \frac{\lambda}{4!}\phi(x)^4 \big)|0>$$
is only one kind of diagram; a disconnected one with no external legs. And that doesn't contribute to any scattering?

 

[olgranpappy]

$$<0|T\big( -i\int d^4x \frac{\lambda}{4!}\phi(x)^4 \big)|0>$$
is only one kind of diagram; a disconnected one with no external legs. And that doesn't contribute to any scattering?

but that doesn't mean it is equal to zero. The disconnected diagrams in a scattering amplitude factor and cancel with the "denominator" of, say <T[phi_x phi_y phi_z phi_w S(inf)]>/<S(inf)>, but they aren't zero, they just don't "contribute" to the amplitude.

if you want to calculate, say, the free energy then you have to evaluate diagram w no external legs explicitly.

 

[kdv]

$$<0|T\big( -i\int d^4x \frac{\lambda}{4!}\phi(x)^4 \big)|0>$$
is only one kind of diagram; a disconnected one with no external legs. And that doesn't contribute to any scattering?

You are right. Is this really the calculation you have to do?

Normally, if you have to calculate the scattering of, say, two particles to two particles, you will need to evaluate not th eexpression you wrote above but rather

$$<0|T\big( -i ~ \phi(x_1) \phi(x_2) \phi(x_3) \phi(x_4) \int d^4x \frac{\lambda}{4!}\phi(x)^4 \big)|0>$$

i.e. there are fields connected to the external spacetime points. Are you sure you don't have those fields as well? If not, then you get only disconnected diagrams as you said. And it makes sense that there is no scattering from your expression since there is no extrenal field to connect to!